
\prob{00C0}{反比例递增}

定义数列
\[ x_n = \begin{cases}
  5, & n = 0 \\ x_{n - 1} + 1/x_{n - 1}, & n > 0
\end{cases} \]
求证：$45 < x_{1000} < 45.1$。
\problabels{yellow/代数, green/证明题}

\emph{ZYT 提供的题目。}

\subsection{分组}

首先，$\forall n > 0$ 有
\begin{align*}
  x_n^2 &= \left(x_{n - 1} + \frac1{x_{n - 1}}\right)^2 = x_{n - 1}^2 + \frac1{x_{n - 1}^2} + 2 \\
  &= x_0^2 + 2n + \sum_{k = 0}^{n - 1} \frac1{x_k^2} > x_0^2 + 2n
\end{align*}
于是
\[ x_{1000}^2 > x_0^2 + 2\cdot1000 = 2025 \Rightarrow x_{1000} > 45 \]
左边不等式得证。

显然数列 $\{x_n\}$ 单调增，故数列 $\left\{1/x_n^2\right\}$ 单调减。于是
\[ \sum_{k = n_0}^{n_1} \frac1{x_k^2} < \frac{n_1 - n_0 + 1}{x_{n_0}^2} \]
故有
\begin{align*}
  \sum_{k = 0}^{24} \frac1{x_k^2} &< \frac{25}{x_0^2} = 1 \\
  \sum_{k = 25}^{99} \frac1{x_k^2} &< \frac{75}{x_{25}^2}, \sum_{k = 100}^{999} \frac1{x_k^2} < \frac{900}{x_{100}^2}
\end{align*}
于是
\begin{align*}
  & x_{25}^2 = x_0^2 + 2\cdot25 + \sum_{k = 0}^{24} \frac1{x_k^2} \\
  \Rightarrow{}& 75 < x_{25}^2 < 76 \Rightarrow \frac1{76} < \frac1{x_{25}^2} < \frac1{75} \\
  \Rightarrow{}& \sum_{k = 25}^{99} \frac1{x_k^2} < \frac{75}{x_{25}^2} < 75\cdot\frac1{75} = 1 \\
  \Rightarrow{}& \sum_{k = 0}^{99} \frac1{x_k^2} < 1 + 1 = 2
\end{align*}
又有
\begin{align*}
  & x_{100}^2 = x_0^2 + 2\cdot100 + \sum_{k = 0}^{99} \frac1{x_k^2} \\
  \Rightarrow{}& 225 < x_{100}^2 < 227 \Rightarrow \frac1{227} < \frac1{x_{100}^2} < \frac1{225} \\
  \Rightarrow{}& \sum_{k = 100}^{999} \frac1{x_k^2} < \frac{900}{x_{100}^2} < 900\cdot\frac1{225} = 4 \\
  \Rightarrow{}& \sum_{k = 0}^{999} \frac1{x_k^2} < 2 + 4 = 6
\end{align*}
因此，有
\begin{align*}
  x_{1000}^2 &= x_0^2 + 2\cdot1000 + \sum_{k = 0}^{999} \frac1{x_k^2} \\
  &< 25 + 2000 + 6 = 2031
\end{align*}
故 $x_{1000} < \sqrt{2031} < 45.1$，右边不等式得证。

证毕。
